[SOLVED] f integrable implies f^2 integrable

**nqramjets** · June 1st 2009, 09:24 PM

I'm trying, unsuccessfully at the moment, to show the following, which I know is true: If $f$ is Riemann integrable, then $f^2$ is Riemann integrable. And, I need to do this without the u,v integrable implies uv integrable. I know there's a way to do it, so if you can point me in the right direction, that would be great! Thanks!

**chisigma** · June 1st 2009, 10:50 PM

A simple example...

$\int_{0}^{1} \frac{dx}{\sqrt{x}}$ exists...

$\int_{0}^{1} \frac{dx}{x}$ doesn't exist...

Kind regards

$\chi$ $\sigma$

**nqramjets** · June 2nd 2009, 08:08 AM

f integrable on [0,1] implies f bounded on [0,1], by contraposition, the integral of 1/sqrt{x} does not exist.

**putnam120** · June 2nd 2009, 08:48 AM

Can you use that if $g(x)$ is continuous and $f(x)$ integrable then $(g\circ f)(x)$ is integrable?

If so then just take $g(x)=x^2$ .

**nqramjets** · June 2nd 2009, 08:51 AM

Yessir, that's the trick. The composition of an integrable function into a continuous function maintains integrability. So, yea, taking your continuous function to be $t^2$ or whatever gives that $f\in R[a,b]\Rightarrow f^2\in R[a,b]$
Cheers!

**chisigma** · June 2nd 2009, 10:04 AM

Originally Posted by nqramjets

f integrable on [0,1] implies f bounded on [0,1], by contraposition, the integral of 1/sqrt{x} does not exist.

I'm afraid I cant agree with you. Starting from the general integration formula...

$\int x^{a}\cdot dx = \frac{x^{1+a}}{1+a} + c$

... where a is an arbitrary real [or even complex...] number provided only that $a \ne -1$ , proceeding in trivial way we have...

$\int_{0}^{1} \frac{dx}{\sqrt{x}} = |2\cdot \sqrt{x}|_{0}^{1}= 2$

Kind regards

$\chi$ $\sigma$

**nqramjets** · June 2nd 2009, 11:04 AM

Okay, this may be a slight difference of definition, but here is the issue:
A function is $f:[0,1]\rightarrow \mathbb{R}$ is Riemann Integrable on $[0,1]$ if there is an $L\in \mathbb{R}$ such that for all $\varepsilon >0$ there exists $\delta >0$ such that for every tagged partition $\dot{P}$ of $[0,1]$ with $\|\dot{P}\|<\delta$ , then
$|S(f;\dot{P})-L|<\varepsilon$
Now, if you look at the Riemann sum part you have
$S(f;\dot{P})=\sum_{i=1}^nf(t_i)(x_i-x_{i-1})$ but if I choose my tag of the first sub-interval to be 0, ie. $t_1=0\in [x_0,x_1]$ then $f(t_i)=\frac{1}{\sqrt{t_i}}=\frac{1}{0}$ which is undefined. So, $f\not\in R[0,1]$ and therefore we cannot apply the FTC.
$\frac{1}{\sqrt{x}}$ simply is not a function from $[0,1]$ into $\mathbb{R}$ because 0 is not in the domain.
What is your definition?

**putnam120** · June 2nd 2009, 11:18 AM

Originally Posted by nqramjets

Okay, this may be a slight difference of definition, but here is the issue:
A function is $f:[0,1]\rightarrow \mathbb{R}$ is Riemann Integrable on $[0,1]$ if there is an $L\in \mathbb{R}$ such that for all $\varepsilon >0$ there exists $\delta >0$ such that for every tagged partition $\dot{P}$ of $[0,1]$ with $\|\dot{P}\|<\delta$ , then
$|S(f;\dot{P})-L|<\varepsilon$
Now, if you look at the Riemann sum part you have
$S(f;\dot{P})=\sum_{i=1}^nf(t_i)(x_i-x_{i-1})$ but if I choose my tag of the first sub-interval to be 0, ie. $t_1=0\in [x_0,x_1]$ then $f(t_i)=\frac{1}{\sqrt{t_i}}=\frac{1}{0}$ which is undefined. So, $f\not\in R[0,1]$ and therefore we cannot apply the FTC.
$\frac{1}{\sqrt{x}}$ simply is not a function from $[0,1]$ into $\mathbb{R}$ because 0 is not in the domain.
What is your definition?

I would also have to agree, since for a function to be Riemann integrable on an interval it has to be bounded. Which $\frac{1}{\sqrt{x}}$ clearly isn't on [0,1]. So even though $\int_0^1\frac{dx}{\sqrt{x}}$ exists, it is actually found by doing the following: $\lim_{c\to 0^+}\int_c^1\frac{dx}{\sqrt{x}}$ .

**nqramjets** · June 2nd 2009, 11:33 AM

Exactly, By definition, 1/sqrt{x} is not Riemann integrable, so the theorem that show f-->f^2 does not apply. Hence this does not break the rules. After all, a limit of an integral is a limit of a limiting process, which is where everything gets alittle funny. So, in conclusion, we are both right: f integrable implies f^2 integrable, but improper integrals do not necessarily obey the same rules.

**Plato** · June 2nd 2009, 12:04 PM

$\chi\sigma$ has confused problems from integration theory with an improper integral from basic calculus.
In basic calculus, $\int_0^1 {\frac{{dx}}{{\sqrt x }}}$ exist and equals $2$ .
But it is also known as an improper integral because $\frac{1}{\sqrt x}$ is not defined at $x=0$ .

In integration theory the function $\frac{1}{\sqrt x}$ is not Riemann integrable on $[0,1]$ , not being bounded there.
But it does have an improper integral there.

**Sampras** · June 2nd 2009, 01:17 PM

Originally Posted by Plato

$\chi\sigma$ has confused problems from integration theory with an improper integral from basic calculus.
In basic calculus, $\int_0^1 {\frac{{dx}}{{\sqrt x }}}$ exist and equals $2$ .
But it is also known as an improper integral because $\frac{1}{\sqrt x}$ is not defined at $x=0$ .

In integration theory the function $\frac{1}{\sqrt x}$ is not Riemann integrable on $[0,1]$ , not being bounded there.
But it does have an improper integral there.

I believe it is Lebesgue integral though.

**nqramjets** · June 2nd 2009, 01:35 PM

Yes, I'm pretty sure it is.

**chisigma** · June 2nd 2009, 11:42 PM

The original Riemann definition is reported here...

http://en.wikipedia.org/wiki/Riemann_integral

... and here we can read...

Given an f(*) defined in the closed interval $[a,b]$ , the finite sequence $a = x_{0}<x_{1}<...<x_{n} = b$ , the finite sequence $t_{0},t_{1},...,t_{n-1}$ subject to the condition $x_{i}\le t_{i} \le x_{i+1}$ , a real number $\epsilon >0$ , if exists a real number $\delta>0$ so that for any partition $\overrightarrow{x},\overrightarrow{t}$ is...

$|\sum_{i=0}^{n-1} f(t_{i})\cdot (x_{i+1}-x_{i}) -s|<\epsilon$ (1)

... we define...

$s=\int_{a}^{b} f(x)\cdot dx$ (2)

Now I have indicated as example the function...

$f(x) = \frac{1}{\sqrt{x}}$ , $0<x\le 1$ (3)

It is obvious that the (3) is not defined in $x=0$ , but, as in most of such cases, we can resolve the abiguity defining f(*) [for example...] as follows...

$f(x)= \frac{1}{\sqrt{x}}$ for $0<x\le 1$ , $=0$ for $x=0$ (4)

... so that is...

$f^{2}(x)= \frac{1}{x}$ for $0<x\le 1$ , $=0$ for $x=0$ (5)

Now we have to verify if $f(*)$ defined in (4) is Riemann integrable according to (1) and if $f^{2}(*)$ defined in (5) is not Riemann integrable according to (1)...

Kind regards

$\chi$ $\sigma$

**chisigma** · June 3rd 2009, 04:00 AM

Now we try to compute the 'Riemann sum' ...

$\sum_{i=0}^{n-1} f(t_{i})\cdot (x_{i+1} - x_{i})$ (1)

... having as goal to arrive to the integral...

$\int_{0}^{1} \frac{dx}{\sqrt{x}} = 2$ (2)

The most simple choice is...

$x_{i}= \frac {i}{n}$

$t_{i}= 0$ for $i=0$ , $t_{i}=\frac{i+\alpha}{n}$ for $i=1,2,...,n-1$ with $0<\alpha<1$ (3)

Defining...

$f(x)= \frac{1}{\sqrt{x}}= 0$ in $x=0$ (4)

... the integral (2) is...

$\int_{0}^{1} \frac{dx}{\sqrt{x}} = lim_{n \rightarrow \infty} \sum_{i=1}^{n-1} \frac {f(t_{i})}{n} = lim_{n\rightarrow \infty} \frac{1}{\sqrt{n}}\cdot \sum _{i=1}^{n-1} \frac{1}{\sqrt{i +\alpha}}$ (5)

The limit (5) has been computed in elegant fashion in ...

http://www.mathhelpforum.com/math-he...-sequence.html

Kind regards

$\chi$ $\sigma$

**nqramjets** · June 8th 2009, 12:15 PM

Obviously if I want to compute the integral of 1/sqrt{x} I can, it can be done is second quarter calculus. The point was that 1/sqrt{x} is NOT Reimann integrable in the strictest, by definition, sense. As a counterexample to f integrable --> f^2 integrable you MUST start with a function that is Reimann integrable. 1/sqrt{x} simply is not on [0,1]. It doesn't matter how you alter the function so that it becomes defined, for it is obviously then not the same function. So the implication is ambiguous.

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